If I understand your question correctly, you can't do better. Indeed from the proof of Erdos-Ko-Rado you can deduce that only the stars have size equal to $\binom{n-1}{r-1}$, when $2r<n$. In the exceptional case $r=n/2$ you can take one of every pair of complementary sets $A,A^c$.
EDIT: The intended question seems to be (equivalent to) the following. How large can an intersecting family of $r$-sets be if we stipulate that it is not a star?
The answer in this case was given by Hilton and Milner (see A. J. W. Hilton and E. C. Milner. Some Intersection Theorems For Systems of Finite Sets Q J Math (1967) 18(1): 369-384 doi:10.1093/qmath/18.1.369).
The largest such family is obtained by fixing an element $1$, an $r$-set $A_1 = \{2,\dots,r+1\}$, and requiring every further set to contain $1$ and to intersect $A_1$. Thus the maximum size is whatever the size of this family is.
Since you seem to be particularly interested in the case of $n$ odd and $r=\lfloor n/2\rfloor$, let's do the calculation in that case. The family you propose ("exclude two points") has size $\binom{n-2}{r} = \binom{n-2}{r-1}$, while the family that Hilton and Milner propose has size $$1 + \binom{n-1}{r-1} - \binom{n-r-1}{r-1} = \binom{n-1}{r-1} - r + 1 = \binom{n-2}{r-1} + \binom{2r-1}{r-2} - r + 1.$$
Sorry, they win. :)
Here is an outline of a proof that seems to work (but there might be more elegant solutions).
1. Notation. Write $[n] = \{1,\ldots,n\}$, and let $\mathcal P([n]) = \{S : S \subseteq [n]\}$ be its power set. I will occasionally refer to the elements of $[n]$ as symbols.
2. Situation. Let $\mathcal S \subseteq \mathcal P([n])$ be a collection of $3$-element sets such that $|T_1 \cap T_2| \neq 1$ for all $T_1,T_2 \in \mathcal S$.
3. Definition. Let $\mathcal S$ be as in Situation 2. Define $G_{\mathcal S} = (V,E)$ to be the graph given by $V = \mathcal S$ and $E = \{\{T_1,T_2\} : T_1 \cap T_2 \neq \varnothing\}$. In other words, $V$ has a vertex for every $3$-element set in $\mathcal S$, and two sets/vertices $T_1,T_2 \in \mathcal S$ are connected if $T_1$ and $T_2$ intersect. (Note: if $T_1$ and $T_2$ are adjacent in $G_{\mathcal S}$, then $|T_1 \cap T_2| = 2$.)
4. Claim. If $\mathcal S$ is as in Situation 2, then $G_{\mathcal S}$ is a disjoint union of cliques.
Proof. Let $T_1T_2\cdots T_k$ be a path in $G_{\mathcal S}$ with $T_1 \neq T_k$. Then $|T_1 \cap T_k|$ is either $0$ or $2$.
Suppose $|T_1 \cap T_k| = 0$. Let $i$ be the least index such that $|T_i \cap T_k| \neq 0$. Then $|T_i \cap T_k| \geq 2$, by the assumption on $\mathcal S$ (cf. Situation 2). But $T_i$ contains only one element (i.e. symbol) which is not contained in $T_{i-1}$, so we also have $|T_{i-1} \cap T_k| > 0$: a contradiction. We conclude that $|T_1 \cap T_k| = 2$, so $T_1$ and $T_k$ are connected by an edge. $\quad\Box$
5. Claim. The only possible cliques in $G_{\mathcal S}$ are of the form $\{\{a,b,c_1\},\{a,b,c_2\},\{a,b,c_3\},\ldots\}$, or a subset of $\{\{a,b,c\},\{a,b,d\},\{a,c,d\},\{b,c,d\}\}$.
Proof. Suppose that $\mathcal C \subseteq \mathcal S$ is a clique in $G_{\mathcal S}$. Choose two subsets $T_1,T_2 \in \mathcal C$, which are necessarily of the form $T_1 = \{a,b,c\}$ and $T_2 = \{a,b,d\}$. Let $T_3 \in \mathcal C$ be distinct from $T_1$ and $T_2$. We distinguish two cases.
If $T_3$ contains both $a$ and $b$, then we claim that $\mathcal C$ is of the form $\{\{a,b,c_1\},\{a,b,c_2\},\{a,b,c_3\},\ldots\}$. Indeed, write $T_3 = \{a,b,e\}$, and let $T_k \in \mathcal C$ be different from $T_1$, $T_2$, and $T_3$. If $T_k$ does not contain $a$, then $|T_1 \cap T_k| = 2$ forces $b,c \in T_k$; $|T_2 \cap T_k| = 2$ forces $b,d \in T_k$; and $|T_3 \cap T_k| = 2$ forces $b,e \in T_k$. But now $|T_k| \geq 4$, which is a contradiction. Analogously, $T_k$ must contain $b$, so every $T_k \in \mathcal C$ contains $a$ and $b$.
If $T_3$ does not contain both $a$ and $b$, then $|T_1 \cap T_3| = |T_2 \cap T_3| = 2$ forces $T_3 = \{a,c,d\}$ or $T_3 = \{b,c,d\}$. By rearranging $a$ and $b$, we may assume without loss of generality that $T_3 = \{a,c,d\}$. Let $T_k \in \mathcal C$ be different from $T_1$, $T_2$, and $T_3$. Then
$$ |T_k \cap \{a,b,c\}| = |T_k \cap \{a,b,d\}| = |T_k \cap \{a,c,d\}| = 2, $$
so the only remaining possibility is $T_k = \{b,c,d\}$. $\quad\Box$
6. Proposition. If $\mathcal S$ is as in Situation 2, then $|\mathcal S| \leq n$.
Proof. For every clique $\mathcal C$ in $G_{\mathcal S}$ one has $|\mathcal C| \leq \big|\bigcup_{T\in\mathcal C} T\big|$, by Claim 5. (In words: every clique uses at least as many symbols $a,b,c,\ldots \in [n]$ as it has vertices.) Since $G_{\mathcal S}$ is a disjoint union of cliques, and since different cliques use different symbols, it follows that $|\mathcal S| \leq n$. $\quad\Box$
Best Answer
HINT: Write as $V$ the $n$-element set. Then there are $2^n$ subsets of $V$, and furthermore those $2^n$ subsets of $V$ can be partitioned into $2^{n-1}$ pairs $\{S,\bar{S}\}$, where $\bar{S}$ is the complement of $S$. If $\cal{F}$ is intersecting however, then for each $S \in \cal{F}$, its complement $\bar{S}$ in $V$ is not in $\cal{F}$, equivalently, only one of $S$, $\bar{S}$ can be in $\cal{F}$.